derivative with quotient rule and summations

Question

I want to differentiate this with respect to $\eta$:

$$C(\eta) = \frac{1+\sum\limits_{k=1}^{3} B_k\left(\frac{\eta}{\eta_c}\right)^k}{3\left(1-\frac{\eta}{\eta_c}\right)}$$

Does this solution look correct? I tried going through with the quotient rule, but it is a bit messy, so I was hoping someone could check my math

$$C'(\eta) = \frac{\left[3\left(1-\frac{\eta}{\eta_c}\right)\sum\limits_{k=1}^{3} B_kk\left(\frac{\eta}{\eta_c}\right)^{k-1}\frac{1}{\eta_c}\right]-\left[\left(1+\sum\limits_{k=1}^{3} B_k\left(\frac{\eta}{\eta_c}\right)^k\right)\left(-\frac{3}{\eta_c}\right)\right]}{\left[3\left(1-\frac{\eta}{\eta_c}\right)\right]^2}$$

Answer 1

You could have made life a bit simpler writing $$C(\eta) = \frac{1+\sum\limits_{k=1}^{3} B_kx^k}{3\left(1-x\right)}\qquad \text{where}\qquad x=\frac{\eta}{\eta_c}$$ and use $$\frac{dC(\eta)}{d\eta} = \frac{d C(\eta)}{dx}\frac{dx}{d\eta}=\frac{1}{\eta_c}\frac{d C(\eta)}{dx}$$ For sure, your result is correct but, since $k$ has only a few values, you could simplity the final result to get $$C'(\eta)=\frac{1}{\eta_c}\frac{1+B_1+2 B_2 x+(3 B_3-B_2) x^2-2 B_3 x^3 }{3(1-x)^2}$$

Answer 2

Here's how you could differentiate this particular function:

$$ C'(\eta) = \left[\frac{1+\sum\limits_{k=1}^{3} B_k\left(\frac{\eta}{\eta_c}\right)^k}{3\left(1-\frac{\eta}{\eta_c}\right)}\right]'= \left[\frac{1+B_{1}\frac{\eta}{\eta_c}+B_{2}\frac{\eta^2}{\eta_c^2}+B_{3}\frac{\eta^3}{\eta_c^3}}{3\left(1-\frac{\eta}{\eta_c}\right)}\right]'=\\ \left[\left(1+B_{1}\frac{\eta}{\eta_c}+B_{2}\frac{\eta^2}{\eta_c^2}+B_{3}\frac{\eta^3}{\eta_c^3}\right)\left[3\left(1-\frac{\eta}{\eta_c}\right)\right]^{-1}\right]'=\\ \left(1+B_{1}\frac{\eta}{\eta_c}+B_{2}\frac{\eta^2}{\eta_c^2}+B_{3}\frac{\eta^3}{\eta_c^3}\right)'\left[3\left(1-\frac{\eta}{\eta_c}\right)\right]^{-1}+\left(1+B_{1}\frac{\eta}{\eta_c}+B_{2}\frac{\eta^2}{\eta_c^2}+B_{3}\frac{\eta^3}{\eta_c^3}\right)\left(\left[3\left(1-\frac{\eta}{\eta_c}\right)\right]^{-1}\right)'=\\ \frac{B_{1}\frac{1}{\eta_c}+B_{2}\frac{2\eta}{\eta_c^2}+B_{3}\frac{3\eta^2}{\eta_c^3}}{3\left(1-\frac{\eta}{\eta_c}\right)}+\left(1+B_{1}\frac{\eta}{\eta_c}+B_{2}\frac{\eta^2}{\eta_c^2}+B_{3}\frac{\eta^3}{\eta_c^3}\right)(-1)\left[3\left(1-\frac{\eta}{\eta_c}\right)\right]^{-2}\left(3-3\frac{\eta}{\eta_c}\right)'=\\ \frac{\sum\limits_{k=1}^{3}B_k\frac{k\eta^{k-1}}{\eta_c^k}}{3\left(1-\frac{\eta}{\eta_c}\right)}+\frac{1+\sum\limits_{k=1}^{3} B_k\left(\frac{\eta}{\eta_c}\right)^k}{\left[3\left(1-\frac{\eta}{\eta_c}\right)\right]^2}\cdot\frac{3}{\eta_c}=\\ \frac{\sum\limits_{k=1}^{3}B_k\frac{k\eta^{k-1}}{\eta_c^k}}{3\left(1-\frac{\eta}{\eta_c}\right)}+\frac{1+\sum\limits_{k=1}^{3} B_k\left(\frac{\eta}{\eta_c}\right)^k}{3\eta_c\left(1-\frac{\eta}{\eta_c}\right)^2}. $$

The answer I get appears to be equivalent to what you got.